hw 34 - nguyen (jmn727) – homework 34 – Turner –...

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Unformatted text preview: nguyen (jmn727) – homework 34 – Turner – (59070) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A vertically pulsed laser fires a 1000 MW pulse of 200 ns duration at a small 10 mg pel- let at rest. The pulse hits the mass squarely in the center of its bottom side. The speed of light is 3 × 10 8 m / s and the acceleration of gravity is 9 . 8 m / s 2 . t h y T b b b b b b b b b b b b b b b b b b b b b b b b v T is the time to reach its maximum height h 10 mg 1000 MW 200 ns If the radiation is completely absorbed without other effects, what is the maximum height the mass reaches? Correct answer: 226 . 757 μ m. Explanation: Let : P = 1000 MW = 1 × 10 9 W , Δ t = 200 ns = 2 × 10- 7 s , m = 10 mg = 1 × 10- 5 kg , L = 4 cm = 4 × 10 6 m , c = 3 × 10 8 m / s , and g = 9 . 8 m / s 2 . Applying conservation of energy, we obtain K f − K i + U f − U i = 0 . Since U i = K f = 0 and K i = p 2 i 2 m , we have − p 2 i 2 m + U f = 0 − p 2 i 2 m + mg h = 0 . (1) Using conservation of momentum, p i = p em = U c = P Δ t c . (2) Substituting p i from Eq. 2 into Eq. 1, we obtain − parenleftbigg P Δ t c parenrightbigg 2 2 m + mg h = 0 − P 2 (Δ t ) 2 2 mc 2 + mg h = 0 . (3) Solving Eq. 3 for the maximum height of the pellet’s trajectory gives us h = P 2 (Δ t ) 2 2 m 2 g 1 c 2 (4) = (1 × 10 9 W) 2 (2 × 10- 7 s) 2 2 (1 × 10- 5 kg) 2 (9 . 8 m / s 2 ) × 1 (3 × 10 8 m / s) 2 = (0 . 000226757 m) (1 × 10 6 μ m / m) = 226 . 757 μ m . 002 10.0 points A thin tungsten filament of length 0 . 591 m radiates 99 . 5 W of power in the form of elec- tromagnetic waves. A perfectly absorbing surface in the form of a hollow cylinder of ra- dius 2 . 37 cm and length 0 . 591 m is placed concentrically with the filament. Assume: The radiation is emitted in the radial direction, and neglect end effects. The speed of light is 2 . 99792 × 10 8 m / s. Calculate the radiation pressure acting on the cylinder. Correct answer: 3 . 77126 × 10- 6 N / m 2 . Explanation: Let : r = 2 . 37 cm = 0 . 0237 m , ℓ = 0 . 591 m , c = 2 . 99792 × 10 8 m / s , and P = 3 . 77126 × 10- 6 N / m 2 . nguyen (jmn727) – homework 34 – Turner – (59070) 2 The intensity of the radiation reaching the walls of the cylinder is I = ( S ) = P 2 π r ℓ , so the radiation pressure on the walls is p = I c = P 2 π r ℓc = 3 . 77126 × 10- 6 N / m 2 2 π (0 . 0237 m) (0 . 591 m) (2 . 99792 × 10 8 m / s) = 3 . 77126 × 10- 6...
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This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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hw 34 - nguyen (jmn727) – homework 34 – Turner –...

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